To evaluate the improper integral \"/2 cos(x) 2 dx, 0 (2 sin(x) 1)? we need to first split the integration at the singular point at x = a, where a = -: \"_ cos x \"/2 cos x / 42 dx + f #2 d3 0 (2 sin(x) 1)? 0+ (2 sin(x) 1)? We then apply the substitution \"(a)_ 1 \"(Jr/2) 1 k/ n dbl + k/ n dbl, u(0) Ll u(a)+ Ll Use this result to determine whether or not the original improper integral converges; if it does, evaluate it; otherwise enter DNE: / \"/2 cos(x) dx _ 0 (2 sin(x) 1)